Question

Two particles of masses `m_(1)` and `m_(2)` initially at rest a infinite distance from each other, move under the action of mutual gravitational pull. Show that at any instant therir relative velocity of approach is `sqrt(2G(m_(1)+m_(2))//R)` where `R` is their separation at that instant.

Answer 1

The gravitational force of attraction on `m_(1)` due to `m_(2)` at a separation r is
`F_(1) = (Gm_(1)m_(2))/(r^(2))`
Therefore the acceleration of `m_(1)` is `a_(1) = (F_(1))/(m_(1)) = (Gm_(2))/(r^(2))`
Similarly the accleration of `m_(2)` due to `m_(1)` is `a_(2) = - (Gm_(1))/(r^(2))`
the negative sign being put as `a_(2)` is directed opposite to `a_(1)`. The relative acclleration of approach is
`a = a_(1) - a_(2) = (G(m_(1) + m_(2)))/(r^(2))`...(1)
If `v` is the relative velocity then `a = (dv)/(dt) = (dv)/(dr) (dr)/(dt)`
`But - (dr)/(dt) =v` (negative sign shows that r decreases with increasing t)
`:. a = - (dv)/(dr) v`...(2)
From (1) and (2) we have
`vdv = - (G(m_(1) + m_(2)))/r^(2)dr`
Integrating we get `(v^(2))/(2) = (G(m_(1)+m_(2)))/(r) + C`
At `r = oo`, v = 0 (given) and so C = 0
`therefore v^(2) = (2G (m_(1)+m_(2)))/(r)`
Let `v = v_(R)` when `r = R` Then `v_(R)=sqrt(((2G(m_(1)+m_(2)))/(R)))` .